Re: [xsl] Sum of identical nodes

["Joris Gillis" <roac@xxxxxxxxxx>]

Hi,

Tempore 08:18:34, die 07/20/2005 AD, hinc in  
xsl-list@xxxxxxxxxxxxxxxxxxxxxx scripsit Alvin Ng <ngkwangming@xxxxxxxxx>:

> Each file in File1 and File2 have few hundreds nodes of <moid></moid>
> and <mt></mt>. My biggest hindrance is how to recursively sum up each
> node from each nodeset.

if the element structures in File1 and File2 are always the same, you can  
use something very simple:

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";  
version="1.0">
<xsl:output method="xml" indent="yes"/>

<xsl:param name="file1" select="/"/>
<xsl:param name="file2" select="document('file2.xml')"/>

<xsl:template match="/">
<xml>
	<xsl:apply-templates select="$file1/xml/*" mode="merge"/>
</xml>
</xsl:template>

<xsl:template match="mt" mode="merge">
	<xsl:variable name="pos" select="position()"/>
	<xsl:copy>
		<xsl:value-of select="sum(. | $file2/xml/*[$pos])"/>
	</xsl:copy>
</xsl:template>

<xsl:template match="moid" mode="merge">
	<xsl:copy-of select="."/>
</xsl:template>

</xsl:stylesheet>


regards,
--
Joris Gillis (http://users.telenet.be/root-jg/me.html)
"N N1N;N.N8N5N9N1 N:N1N9 ON? N;N,N4N9 ON,N=ON1 N2N3N1N/N=N?ON= N1OO ON,N=O	"